A small disc of mass m slides down with initial velocity zero from the top (A) of a smooth hill of height H having a horizontal portion (B C) as shown in the figure. If the height of the horizontal portion of the hill is h , then the maximum horizontal distance covered by the disc from the point D is

Step-by-step solution

According to law of conservation of energy at point B shown in the figure given in question part, Loss in PE = Gain in KE m g(H-h)= 1 2 m v^2 v= 2 g(H-h) Now, h= 1 2 g t^2 t= 2 h g Distance covered in horizontal portion, s=v t= 2 g(H-h) 2 h g s= 4 h(H-h) ...(i) For maximum value of s d s d h =0 2(H-2 h)=0 H=2 h h= H 2 Substituting the value of h in Eq. (i), we get, aligned s & = 4 H 2 (H- H 2 ) & = 2 H H 2 =H aligned